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SpyIntel [72]
2 years ago
10

How do i Write in Digits the fifty five hundredths​

Mathematics
2 answers:
grandymaker [24]2 years ago
6 0

fifty five hundredths is equal to 0.55

stellarik [79]2 years ago
6 0
Veihcjssjsjjssjsbudhdisjidisjssjssskdjjdkd
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With a little care, you can find the median and the quartiles from the histogram. what are these numbers? you can also find the
coldgirl [10]

Complete Question

The histogram from this question is shown on the first uploaded image

Answer:

The first quartile is  1st Q = 19^{th} girl (subject ) = 1 \serving\ of\ fruit\ per\ day

The median is   Median  = 38 ^{th} \ girl (subject) =  2 \serving\ of\ fruit\ per\ day

The third quartile  is 3rd Q =  56^{th} \  girl (subject) = 4 \serving\ of\ fruit\ per\ day  

The  mean is  \= x = 2.62

Step-by-step explanation:

From the question we are told that

   The total  number of girls is n =  74

Generally the median is mathematically represented as

       Median  =  \frac{\frac{n}{2}  +[ \frac{n}{2} + 1]  }{2}

So

          Median  =  \frac{\frac{74}{2}  +[ \frac{74}{2} + 1]  }{2}

=>         Median  =  \frac{37  +38 }{2}

=>         Median  =37.5 \  girl    

From the histogram 37.5 \approx 38^{th} \ girl fall under 2 fruits per day

Generally the first quartile is mathematically represented as

      1st \ Q =  \frac{\frac{n}{4} + [\frac{n}{4} + 1]  }{2}

=>    1st \ Q =  \frac{\frac{74}{4} + [\frac{74}{4} + 1]  }{2}

=>     1st \ Q = 19 \  girl

From the histogram 19^{th}girl fall under 1 fruit per day

 Generally the third quartile is mathematically represented as

     3rd\  Q =  \frac{\frac{n}{2}  + n }{2}

=>    3rd\  Q =  \frac{\frac{74}{2}  + 74 }{2}

=>    3rd\  Q =  55.5

From the histogram 55.5 \approx 56^{th} \ girl fall under 4 fruits per day

Generally from the histogram table the frequency  [number of subjects (girls) ]  and the servings of fruit per day can be represented as

                                                                                                               Total

x(servings per day )        0     1     2     3     4     5    6     7      8    

f (number of subjects )   15    11   15     11    8     5    3     3      3    \sum f  =  74

xf                                       0    11    30   33   32  25   18   21     24    \sum xf =  194

Generally the mean is mathematically represented as

      \= x = \frac{1}{ \sum f}  * [\sum xf]

=>     \= x = \frac{1}{ 74}  *194

=>      \= x = 2.62

4 0
3 years ago
Evaluate √196 <br>A.11<br>B.12<br>C.13<br>D.14
ANTONII [103]

Hello :)

196 = 14.14 so the answer is D

169 = 13.13

144=12.12

121=11.11

Have a nice day :)

5 0
3 years ago
Read 2 more answers
UVOISINOULLIV HU
Gala2k [10]
<h3>Total snow fall for three days is 8.75 inches</h3>

<em><u>Solution:</u></em>

Given that,

Snow fell over the past three days

First\ day = 3\frac{2}{4}\ inches = \frac{14}{4}\ inches\\\\Second\ day = 3\frac{2}{4}\ inches = \frac{14}{4}\ inches\\\\Third\ day = 1\frac{3}{4}\ inches = \frac{7}{4}\ inches

<em><u>What was the total snowfall for the three days?</u></em>

Total snowfall = first day + second day + third day

Total\ snowfall = \frac{14}{4} + \frac{14}{4} + \frac{7}{4}\\\\Total\ snowfall = \frac{14 + 14 + 7}{4}\\\\Total\ snowfall = \frac{35}{4} \text{ or } 8.75\ inches

Thus total snow fall for three days is 8.75 inches

5 0
3 years ago
X/2 - y/3 = 3/2<br>X/3 + y/2 = 16/3
Ivenika [448]
X/2-y/3=3/2
(6×x/2)-(6×y/3)=6×3/2
3x-2y=9______(1)

x/3+y/2=16/3
(6×x/3)+(6×y/2)=6×16/3
2x+3y=32_____(2)
(1)×3____9x-6y=27____(3)
(2)×2____4x+6y=64____(4)
(3)+(4)___13x=91
x=7
3(7)-2y=9
-2y=-12
y=6
4 0
3 years ago
A classmate said that line AB and line DC are parallel based on the diagram below. Explain your classmates error.
MatroZZZ [7]

<em><u>Explanation</u></em>:

The only information we have to go on is that ∠AWZ ≅ ∠XYC. However, these are not specific kinds of angles that can prove lines parallel. If we had a pair of alternate interior angles, such as ∠AWZ and ∠WZY, we could say that the lines are parallel. However, we do not have enough information.

3 0
3 years ago
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